Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor…

Q25 jee_main_2025_02_april_morning Motion in a Straight Line

A person travelling on a straight line moves with a uniform velocity

v₁

for a distance

x

and with a uniform velocity

v₂

for the next

(3)/(2)x

distance. The average velocity in this motion is

(50)/(7)~m/s

. If

v₁

is

5~m/s

then

v₂ =

____

m/s

.

Numerical Answer. Answer: 10 to 10

Solution

Related Formula

vavg = Total DistanceTotal Time

Core Logic

Let's find the time taken for each section of the motion:

First section of distance $x$ $x$ with velocity $v₁ = 5~m/s$ $v_1 = 5\mathrm{~m/s}$:

t₁ = (x)/(v₁) = (x)/(5)

Second section of distance $(3)/(2)x$ $\frac{3}{2}x$ with velocity $v₂$ $v_2$:

t₂ = (3x/2)/(v₂) = (3x)/(2v₂)

Total distance is:

dtotal = x + (3)/(2)x = (5)/(2)x

Average velocity is:

vavg = dtotalt₁ + t₂ = ((5)/(2)x)/((x)/(5) + (3x)/(2v₂))

We are given $vavg = (50)/(7)~m/s$ . Equating the two values (noting $x$ $x$ cancels out):

(50)/(7) = ((5)/(2))/((1)/(5) + (3)/(2v₂))

Divide both sides by 5:

(10)/(7) = ((1)/(2))/((1)/(5) + (3)/(2v₂)) 10 ((1)/(5) + (3)/(2v₂)) = (7)/(2)

2 + (15)/(v₂) = 3.5 (15)/(v₂) = 1.5 v₂ = (15)/(1.5) = 10~m/s

Step 1: Final Conclusion

The velocity $v₂$ $v_2$ is $10~m/s$ $10\mathrm{~m/s}$.

Pattern Recognition

Never take simple arithmetic averages of velocities! Average velocity must always be calculated as $Total DistanceTotal Time$ . Because total distance and time intervals are proportional to $x$ $x$, $x$ $x$ cleanly cancels out.

Chapter Mix

Class 11 Physics: Kinematics

Q11 jee_main_2025_03_april_evening Projectile Motion

A particle is projected with velocity

u

so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as

(nu²)/(25g)

, where value of

n

is: (Given '

g

' is the acceleration due to gravity).

A. 6
B. 18
C. 12
D. 24

Solution

Related Formula

For a projectile with launch speed $u$ $u$ and angle $θ$ $\theta$:

Horizontal Range:

R = (u² (2θ))/(g) = (2 u² θ θ)/(g)

Maximum Height:

H = (u² ²θ)/(2g)

The general ratio linking range and maximum height is:

θ = (4H)/(R)

Core Logic

Given state:

R = 3H \Rightarrow (H)/(R) = (1)/(3)

Step 1: Determine the projection angle ( $θ$ $\theta$)

Substitute the ratio into the relation:

θ = 4 ((H)/(R)) = 4 ((1)/(3)) = (4)/(3)

This is a standard Pythagorean triangle angle:

θ = (4)/(5), θ = (3)/(5)

Step 2: Compute the horizontal range ( $R$ $R$)

R = (2 u² θ θ)/(g)

R = (2 u² ((4)/(5)) ((3)/(5)))/(g) = (24 u²)/(25 g)

Comparing this with the given format $(nu²)/(25g)$ $\frac{nu^2}{25g}$:

$n = 24$ $n = 24$

Pattern Recognition

The relation $θ = 4H/R$ $\tan\theta = 4H/R$ is an essential identity in projectile dynamics. Whenever $R = k H$ $R = k H$, then $θ = 4/k$ $\tan\theta = 4/k$. Recognizing standard angles like $θ = 4/3$ $\tan\theta = 4/3$ or $3/4$ $3/4$ directly yields trigonometric values immediately.

Chapter Mix

Class 11 Physics: Motion in a Plane

Q14 jee_main_2025_03_april_evening Kinematics and Derivative Relations

A particle moves along the x-axis and has its displacement x varying with time t according to the equation

x=c₀(t²-2)+c(t-2)²

where

c₀

and c are constants of appropriate dimensions. Then, which of the following statements is correct?

A. the acceleration of the particle is $2c₀$ $2c_{0}$
B. the acceleration of the particle is 2c
C. the initial velocity of the particle is 4c
D. the acceleration of the particle is $2(c+c₀)$ $2(c+c_{0})$

Solution

Related Formula

In rectilinear kinematics:

Velocity:

v = (dx)/(dt)

Acceleration:

a = (dv)/(dt) = (d²x)/(dt²)

Core Logic Given position-time function:

x(t) = c₀ (t² - 2) + c (t - 2)²

Step 1: Differentiate once to get velocity ($ Step 1: Differentiate once to get velocity ($v) $) $

v = (dx)/(dt) = (d)/(dt)[c₀(t² - 2)] + (d)/(dt)[c(t-2)²]

v = c₀ (2t) + c \cdot 2(t-2) = 2 c₀ t + 2 c(t - 2)

Step 2: Differentiate again to get acceleration ($ Step 2: Differentiate again to get acceleration ($a) $) $

a = (dv)/(dt) = (d)/(dt)[2 c₀ t + 2 c(t - 2)]

a = 2 c₀ + 2 c = 2(c + c₀)

This shows acceleration is constant and equals$ This shows acceleration is constant and equals $2(c + c_0), matching Statement (4). Pattern Recognition Whenever a position function is a pure quadratic polynomial in$, matching Statement (4). Pattern Recognition Whenever a position function is a pure quadratic polynomial in $t, the acceleration is constant and equal to$, the acceleration is constant and equal to $2 \timesthe coefficient of the$ the coefficient of the $t^2term. Rewriting$ term. Rewriting $x(t):

x(t) = (c₀ + c)t² - 4ct + (4c - 2c₀)

The coefficient of $ The coefficient of $t^2 is $ is $ (c_0 + c) . Thus, acceleration is $. Thus, acceleration is $ 2(c_0 + c)$ directly. Chapter Mix Class 11 Physics: Motion in a Straight Line

Q12 jee_main_2025_07_april_morning Projectile Motion

Two projectiles are fired from ground with same initial speeds from same point at angles

(45° + α)

and

(45° - α)

with horizontal direction. The ratio of their times of flights is

A. 1
B. $(1 - α)/(1 + α)$ $\frac{1 - \tan\alpha}{1 + \tan\alpha}$
C. $(1 + 2α)/(1 - 2α)$
D. $(1 + α)/(1 - α)$ $\frac{1 + \tan\alpha}{1 - \tan\alpha}$

Solution

Related Formula

The time of flight $T$ $T$ of a projectile launched with speed $u$ $u$ at an angle $θ$ $\theta$ with the horizontal is:

T = (2u θ)/(g)

Core Logic The launch angles of the two projectiles are:

θ₁ = 45^° + α

θ₂ = 45^° - α

Since they have the same speed $ Since they have the same speed $u :

(T₁)/(T₂) = ( (45^° + α))/( (45^° - α))

Step 1: Simplify Trigonometric Ratio Using the angle sum and difference formulas:

(T₁)/(T₂) = ( 45^° α + 45^° α)/( 45^° α - 45^° α)

(T₁)/(T₂) = 1\sqrt(2) α + 1\sqrt(2) α 1\sqrt(2) α - 1\sqrt(2) α = ( α + α)/( α - α)

Divide numerator and denominator by$ Divide numerator and denominator by $\cos\alpha:

(T₁)/(T₂) = (1 + α)/(1 - α)

$

Pattern Recognition Sees: Projectile angles complementary to$ Pattern Recognition Sees: Projectile angles complementary to $45^\circ. Shortcut: Remember the identity$. Shortcut: Remember the identity $\tan(45^\circ + \alpha) = \frac{1+\tan\alpha}{1-\tan\alpha}. Since complementary angles have sine ratios proportional to$. Since complementary angles have sine ratios proportional to $\sin(45^\circ + \alpha)/\sin(45^\circ - \alpha) = \tan(45^\circ + \alpha), the answer is directly$, the answer is directly $\frac{1+\tan\alpha}{1-\tan\alpha}$. Chapter Mix Class 11 Physics: Motion in a Plane

Q16 jee_main_2025_08_april_evening Projectile Motion

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is

8

times higher than that of the second ball.

T₁

and

T₂

are the total flying times of first and second ball, respectively, then the ratio of

T₁

and

T₂

is:

A. $2\sqrt(2) : 1$ $2\sqrt{2} : 1$
B. $2 : 1$ $2 : 1$
C. $\sqrt(2) : 1$ $\sqrt{2} : 1$
D. $4 : 1$ $4 : 1$

Solution

Related Formula

H = (u² ²θ)/(2g) and T = (2u θ)/(g)

where, $H$ $H$ = maximum height reached $T$ $T$ = total time of flight $u$ $u$ = initial projection velocity $θ$ $\theta$ = angle of projection

Core Logic

From the formulas, we see that:

H \propto ²θ and T \propto θ

Thus, we can directly link time of flight to the square root of the maximum height:

T \propto \sqrt(H) (T₁)/(T₂) = \sqrt((H₁)/(H₂))

Step 1: Compute Ratio

Given:

H₁ = 8 H₂ (H₁)/(H₂) = 8

Substitute this ratio:

(T₁)/(T₂) = \sqrt(8) = 2\sqrt(2)

Thus, the ratio is $2\sqrt(2) : 1$ $2\sqrt{2} : 1$.

Pattern Recognition

Sees: Projectile heights ratio → Time of flight ratio. Shortcut: Since $H \propto uy²$ $H \propto u_y^2$ and $T \propto uy$ $T \propto u_y$, we have $T \propto \sqrt(H)$ $T \propto \sqrt{H}$. If the height is 8 times larger, the flying time is $\sqrt(8) = 2\sqrt(2)$ $\sqrt{8} = 2\sqrt{2}$ times larger. ✓

Chapter Mix

Class 11 Physics: Kinematics

Solution & Explanation

Related Formula

Core Logic

Step 1: Total Time of Fall

Step 2: Time Interval Between Drops

Step 3: Distance of the Fourth Drop

Step 4: Height from Ground

Pattern Recognition

Chapter Mix

Reference Study Guides

More Motion in a Straight Line Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine the projection angle (θ$\theta$)

Step 2: Compute the horizontal range (R$R$)

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Differentiate once to get velocity ($ Step 1: Differentiate once to get velocity ($v)$)

Step 1: Differentiate once to get velocity ($v)

Step 2: Differentiate again to get acceleration ($ Step 2: Differentiate again to get acceleration ($a)$)

Step 2: Differentiate again to get acceleration ($a)

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Core Logic

Step 1: Simplify Trigonometric Ratio

Step 1: Simplify Trigonometric Ratio

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Ratio

Pattern Recognition

Chapter Mix

More Motion in a Straight Line Questions — jee_main_2026_28_january_morning

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Step 1: Determine the projection angle ( $θ$ $\theta$)

Step 2: Compute the horizontal range ( $R$ $R$)

Step 1: Differentiate once to get velocity ($)

Step 1: Differentiate once to get velocity ($v $)$

Step 2: Differentiate again to get acceleration ($)

Step 2: Differentiate again to get acceleration ($a $)$